How can I split a string in Java and retain the delimiters?

Question

I have this string (Java 1.5):

:alpha;beta:gamma;delta

I need to get an array:

{\":alpha\", \";beta\", \":gamma\", \";delta         


        

Answer 1

This should work with Java 1.5 (Pattern.quote was introduced in Java 1.5).

// Split the string on delimiter, but don't delete the delimiter
private String[] splitStringOnDelimiter(String text, String delimiter, String safeSequence){
    // A temporary delimiter must be added as Java split method deletes the delimiter

    // for safeSequence use something that doesn't occur in your texts 
    text=text.replaceAll(Pattern.quote(delimiter), safeSequence+delimiter);
    return text.split(Pattern.quote(safeSequence));
}

If first element is the problem:

private String[] splitStringOnDelimiter(String text, String delimiter, String safeSequence){
    text=text.replaceAll(Pattern.quote(delimiter), safeSequence+delimiter);
    String[] tempArray = text.split(Pattern.quote(safeSequence));
    String[] returnArray = new String[tempArray.length-1];
    System.arraycopy(tempArray, 1, returnArray, 0, returnArray.length);
    return returnArray;
}

E.g., here "a" is the delimiter:

splitStringOnDelimiter("-asd-asd-g----10-9asdas jadd", "a", "<>")

You get this:

1.: -
2.: asd-
3.: asd-g----10-9
4.: asd
5.: as j
6.: add

If you in fact want this:

1.: -a
2.: sd-a
3.: sd-g----10-9a
4.: sda
5.: s ja
6.: dd

You switch:

safeSequence+delimiter

with

delimiter+safeSequence

Answer 2

/**
 * @param list an empty String list. used for internal purpose. 
 * @param str  String which has to be processed.
 * @return Splited String Array with delimiters.
 */
public  String[] split(ArrayList<String> list, String str){
  for(int i = str.length()-1 ; i >=0 ; i--){
     if(!Character.isLetterOrDigit((str.charAt(i)))) {
        list.add(str.substring(i, str.length()));
        split(list,str.substring(0,i));
        break;
     }
  }
  return list.toArray(new String[list.size()]);
}

Answer 3

You can do this by simply using patterns and matcher class in java regx.

    public static String[] mysplit(String text)
    {
     List<String> s = new ArrayList<String>();
     Matcher m = Pattern.compile("(:|;)\\w+").matcher(text);
     while(m.find()) {
   s.add(m.group());
     }
     return s.toArray(new String[s.size()]);
    }

Answer 4

Assuming that you only have a finite set of seperators before the words in your string (eg ;, : etc) you can use the following technique. (apologies for any syntax errors, but its been a while since I used Java)

String toSplit = ":alpha;beta:gamma;delta "
toSplit = toSplit.replace(":", "~:")
toSplit = toSplit.replace(";", "~;")
//repeat for all you possible seperators
String[] splitStrings = toSplit.split("~")

Answer 5

str.split("(?=[:;])")

This will give you the desired array, only with an empty first item. And:

str.split("(?=\\b[:;])")

This will give the array without the empty first item.

• The key here is the (?=X) which is a zero-width positive lookahead (non-capturing construct) (see regex pattern docs).
• [:;] means "either ; or :"
• \b is word-boundary - it's there in order not to consider the first : as delimiter (since it is the beginning of the sequence)

Answer 6

To keep the separators, you can use a StringTokenizer:

new StringTokenizer(":alpha;beta:gamma;delta", ":;", true)

That would yield the separators as tokens.

To have them as part of your tokens, you could use String#split with lookahead.